Question

Based on the info in the book, I am not sure what is being asked in this problem:   [There are 300 welders employed at the Tsunami Shipyards Corporation; a sample who graduated from BJ's Welding School revealed a welding certification score of 40 with a σ = 6 & a standard error of the mean of 1.3. A 94% interval estimate shows the margin of error of their scores to be ± ? ] Are they looking for E? where E = z (SD / sqrt of n) ?? So far I am taking the 94 % interval divided by 2 ---> .94 / 2 = .47 nearest z value = 1.87

are they looking for E = Z * ( SD /sqrt N) = (1.87) * (6/ sqrt N) , N = (6/1.3) squared??   ---> that doesnt seem right, would yield a sample value n = 21.3

Answer 1

We are here estimating the margin of error not sample size.

And sample size of 21 is not wrong.we cannot say that sample size would not be 21..it can be 21.

Out of 300 size we have get sample of 21 then it would be ok.